# Identity on covariant derivative

I'm looking to prove the following identity for my homework:

"Denoting the Riemann curvature tensor of the first kind by $R^i_{j k l}$ and letting $\left( V^i \right)$ be a contravariant vector, prove that

$V^i_{,k l} - V^i_{, l k} = -R^i_{ r k l} V^r$."

I have been given by my lecturer the following definition:

"Let $T$ be a contravariant vector field, $T = \left( T_i \right)$ in $\left( x^i \right)$. Then the covariant derivative of $T$ w.r.t. $\left(x^k \right)$ is given by

$T_{, k} = \frac{\partial T^i}{\partial x^k} + \Gamma^i_{t k} T^t$."

I've also been given the following,

"In a coordinate system $\left(x^k \right)$ the covariant derivative wrt $x^k$ of a tensor $T = \left( T^{i_1 \ldots i_p}_{j_1 \ldots j_q} \right)$ is $T_{, k} = \left( T^{i_1 \ldots i_p}_{j_1 \ldots j_{q,k}} \right)$ and

$T^{i_1 \ldots i_p}_{j_1 \ldots j_{q,k}} = \frac{\partial T^{i_1 \ldots i_p}_{j_1 \ldots j_q}}{\partial x^k} + \Gamma^{i_1}_{t k} T^{t i_2 \ldots i_p}_{j_1 \ldots j_q} + \Gamma^{i_2}_{t k} T^{i_1 t i_3 \ldots i_p}_{j_1 \ldots j_q} + \cdots + \Gamma^{i_p}_{t k} T^{i_1 i_2 \ldots i_{p-1} t}_{j_1 \ldots j_q} - \Gamma_{j_1 k}^t T^{i_1 \ldots i_p}_{t j_2 \ldots j_q} - \Gamma^t_{j_2 k} T^{i_1 \ldots i_p}_{j_1 t \ldots j_q} - \cdots - \Gamma^t_{j_q k} T^{i_1 \ldots i_p} _{j_1 j_2 \ldots j_{q-1} t} \quad (*)$."

If I use the first of the formulae, I come up with

$V^i_{, kl} = \frac{\partial \left( V^i_{,k} \right)}{\partial x^l} + \Gamma^i_{tl} V^{it} + \Gamma^i_{tk} V^{it}$

and if I use the second, $(*)$, I come up with

$V^i_{,k l} = \frac{\partial \left(V^i_{, k} \right)}{\partial x^l} + \Gamma^i_{t l} V^t_k - \Gamma^t_{k l} V^i_t$. (See comment for correction to this suggestion).

Which, if either, is correct?

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That's non-standard notation; usually a comma is used for the ordinary partial derivative, and a semi-colon for the covariant derivative. – Rhys Nov 22 '12 at 14:37
Oh, I was following Schaum's "Tensor Calculus" where commas are used. – Harry Williams Nov 25 '12 at 1:12

$V^t_k$ and $V^{it}$ are not defined, since $V^i$ is a vector field.
Since the covariant derivative of $V^i$ is a (1,1) tensor, have to use the second formula to do the second derivative.
However, your formula is missing a index. It should be $$T^i_{,k} = \partial_k T^i + \Gamma^i_{kj}T^j.$$ (Didn't check the second one)
I understand now that my first suggestion is certainly incorrect, for $V^{it}$ is not defined. I don't think my formula is missing an index, however, for it was stated that $T = \left(T_i \right)$, no? :) I believe the second one is correct, however, it should read $V^i_{,k l} = \frac{\partial \left(V^i_{, k} \right)}{\partial x^l} + \Gamma^i_{t l} V^t_{,k} - \Gamma^t_{k l} V^i_{,t}$. Any thoughts on this? – Harry Williams Nov 25 '12 at 1:08